Let $K$ be an ordered field containing the field of rationals $\mathbb{Q}$. Prove that $\mathbb{Q}$ is dense in $K$ if and only if every element of $K$ is the limit of a Cauchy sequence in $\mathbb{Q}$.
Definition: $\mathbb{Q}$ is dense in $K$ if for each pair $x,y$ in $K$, $x<y$, there exists $z\in\mathbb{Q}$ such that $x<z<y$.
The $\Leftarrow$ is trivial since for each $k\in K$, $k$ is the limit of a Cauchy sequence in $\mathbb{Q}$, so $k\in\mathbb{Q}\Rightarrow K=\mathbb{Q}$.
I'm stuck at $\Rightarrow$. For $k\in K$, I guess I have to construct a Cauchy sequence in $\mathbb{Q}$ that converges to $k$ by using the density of $\mathbb{Q}$ in $K$. But I don't know how to proceed it. Can someone help me? Thanks in advance.
I'm assuming the ordinate field has a metric structure endowed by its absolute value, i.e. the metric which generates the order topology.
Let $k \in K$. As $K$ is a field then $$ k-\frac{1}{n} \in K $$ By density there exists $q_n \in \mathbb{Q}$ such that $k-\frac{1}{n} < q_n < k-\frac{1}{n+1}$
We have $$ \lim_{n \to \infty} q_n= k $$ Indeed, for any $\epsilon \in K$, $\epsilon>0$ there exists $N$ such that, for each $n>N$ $$ |q_n-k|< \epsilon $$ (see the edit below)
As convergent sequences are Cauchy that means that $k$ is the limit of a Cauchy sequence
Edit:
As said in the comments to show convergence we need to show that for any $h>0 \in K$ there exists $n \in \mathbb{N}/ \{0\}$ such that $$ \frac{1}{n}<h $$ (this is actually a characterization of the Archimedean property for ordered fields ), but as $0 \in K$, by density there exists $q \in \mathbb{Q}$ such that $$ 0<q<h $$ By the Archimedean property of $\mathbb{Q}$ there exists $n$ such that $$ \frac{1}{n}<q<h $$